Source of the "$\cosh$ trick" for Laplacian eigenfunctions or Helmholtz equation solutions?

Solution 1:

Here is a perspective on this method. Suppose we treat $k$ itself as a coordinate, and perform a Fourier transform with respect to it. This converts the Helmholtz equation $(\Delta+k^2)f=0$ to D'Alembert's equation $(\Delta-\partial_0^2)g=0$. (I have taken $x_0$ to be the conjugate of $k$, and $g$ to be the corresponding Fourier transform of $f$.) Upon defining $x_{n+1}\equiv i x_0$, this becomes $(\Delta+\partial_{n+1}^2)g=0$ which is equivalent to the form identified above.

Separating the variable $x_{n+1}$ by writing $g(\mathbf{x},x_{n+1})=f(\mathbf{x})\Phi(x_{n+1})$ then obtains $$\frac{\Delta f}{f}=-\frac{\Phi''}{\Phi}=-k^2$$ where the separation constant is chosen so as to regain the Helmholtz equation. This yields $\Phi(x_{n+1})=A\cosh(kx_{n+1})+B\sinh(kx_{n+1})$, generalizing the case of $\Phi(x_{n+1})=\cosh(kx_{n+1})$ given in the OP.